Question

Hello Mathematicians!
Does anyone knows how to solve it?
\sin \frac{23\pi}{12}

Answer 1

couldnot read it..

Answer 2

@Gintare kindly type it once again in the comment section

Answer 3

\[\sin \frac{ 23\pi }{ 12}\]

Answer 4

\[\frac{23 \pi }{12}=\frac{23 (180 {}^{\circ})}{12}=345 {}^{\circ} \]

Answer 5

Yes, but I need to make reduction. So I get sin(270+75), then -cos(30+45)
cos(a+b)=cosa*cosb-sina*sinb and the problem is that I cannot make the calculation rigth...

Answer 6

Writing \(\dfrac{23\pi}{12}=\dfrac{(22+1)\pi}{12}\), you have\[\begin{align*}\sin\frac{23\pi}{12}&=\sin\frac{11\pi}{6}\cos\frac{\pi}{12}+\sin\frac{\pi}{12}\cos\frac{11\pi}{6}\\[1ex]
&=-\sin\frac{\pi}{6}\cos\frac{\pi}{12}+\sin\frac{\pi}{12}\cos\frac{\pi}{6}\end{align*}\]Write \(\dfrac{\pi}{12}=\dfrac{\pi}{3}-\dfrac{\pi}{4}\), then
\[\begin{align*}\color{white}{\sin\frac{23\pi}{12}}&=-\sin\frac{\pi}{6}\cos\left(\frac{\pi}{3}-\frac{\pi}{4}\right)+\sin\left(\frac{\pi}{3}-\frac{\pi}{4}\right)\cos\frac{\pi}{6}\\[2ex]
&=-\sin\frac{\pi}{6}\left(\cos\frac{\pi}{3}\cos\frac{\pi}{4}+\sin\frac{\pi}{3}\sin\frac{\pi}{4}\right)\\[1ex]
&\quad\quad+\cos\frac{\pi}{6}\left(\sin\frac{\pi}{3}\cos\frac{\pi}{4}-\sin\frac{\pi}{4}\cos\frac{\pi}{3}\right)\end{align*}\]

Answer 7

It's easier (for me) to use degrees
so
sin(345)
we can use -15 instead of 345
sin(345)= sin(-15)
and -15 = 30-45
so
sin(30-45) = sin(30) cos(45) - cos(30) sin(45)
\[ \sin(345)= \frac{1}{2} \frac{\sqrt{2}}{2} - \frac{\sqrt{3}}{2} \frac{\sqrt{2}}{2} \\
= \frac{\sqrt{2}}{4} \left(1 - \sqrt{3}\right)
\]

Answer 8

Thank you very much!!!!